Documents

YearFilenameLanguageSource
1988IMO-1988-problems-eng.pdfen
Problem 1

Consider two coplanar circles of radii RR and rr (R>rR > r) with the same center. Let PP be a fixed point on the smaller circle and BB a variable point on the larger circle. The line BPBP meets the larger circle again at CC. The perpendicular ll to BPBP at PP meets the smaller circle again at AA. (If ll is tangent to the circle at PP then A=PA = P.)

(i) Find the set of values of BC2+CA2+AB2BC^2 + CA^2 + AB^2.

(ii) Find the locus of the midpoint of BCBC.

Problem 2

Let nn be a positive integer and let A1A_1, A2A_2, \ldots, A2n+1A_{2n+1} be subsets of a set BB. Suppose that

(a) Each AiA_i has exactly 2n2n elements,

(b) Each AiAjA_i \cap A_j (1i<j2n+11 \leq i < j \leq 2n + 1) contains exactly one element, and

(c) Every element of BB belongs to at least two of the AiA_i.

For which values of nn can one assign to every element of BB one of the numbers 00 and 11 in such a way that AiA_i has 00 assigned to exactly nn of its elements?

Problem 3

A function ff is defined on the positive integers by

f(1)=1,f(3)=3,f(2n)=f(n),f(4n+1)=2f(2n+1)f(n),f(4n+3)=3f(2n+1)2f(n),\begin{aligned} f(1) &= 1, \quad f(3) = 3, \\ f(2n) &= f(n), \\ f(4n + 1) &= 2f(2n + 1) - f(n), \\ f(4n + 3) &= 3f(2n + 1) - 2f(n), \end{aligned}

for all positive integers nn.

Determine the number of positive integers nn, less than or equal to 1988, for which f(n)=nf(n) = n.

Problem 4

Show that set of real numbers xx which satisfy the inequality

k=170kxk54\sum_{k=1}^{70} \frac{k}{x-k} \geq \frac{5}{4}

is a union of disjoint intervals, the sum of whose lengths is 1988.

Problem 5

ABCABC is a triangle right-angled at AA, and DD is the foot of the altitude from AA. The straight line joining the incenters of the triangles ABDABD, ACDACD intersects the sides ABAB, ACAC at the points KK, LL respectively. SS and TT denote the areas of the triangles ABCABC and AKLAKL respectively. Show that S2TS \geq 2T.

Problem 6

Let aa and bb be positive integers such that ab+1ab + 1 divides a2+b2a^2 + b^2. Show that

a2+b2ab+1\frac{a^2 + b^2}{ab + 1}

is the square of an integer.